If a^2 - 9/12a = a - 3/a + 3 , a + 3 # 0, and a # 0, then a =?

Question

If  \frac{a^2 -9}{12a} =   \frac{a - 3}{a + 3} , a + 3  # 0, and a # 0, then  a = ?

(A) 1
(B) 2
(C) 3
(D) 6
(E) 9

Source: Nova GMAT

Mohammadmo · Answer

(a-3)(a+3)/12a=(a-3)/(a+3)
==》(a+3)^2=12a
==》 a^2+6a+9=12a
==》 a^2-6a+9=0
(a-3)^2=0
==》a=3
Option (c)

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Poorvi05 · Answer

a^2-9 can be written as (a+3)(a-3)
Therefore, we can cancel out (a-3) from both the sides.
.
.
.
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By cross multiplication we get,
(a+3)^2=12a
On further solving we get,
(a-3)(a-3)=0
Hence, a=3

IMO-C

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Archit3110 · Answer

simplify for given expression  \frac{a^2 -9}{12a} =   \frac{a - 3}{a + 3}
we get a^2-6a+9=0
a= 3
IMO C

ScottTargetTestPrep · Answer

Cross multiplying, we have:

(a + 3)(a - 3)(a + 3) = 12a(a - 3)

(a - 3)(a + 3)^2 - 12a(a - 3) = 0

(a - 3)  = 0

At this point, we see that if a - 3 = 0, or a = 3, the equation will be true.

Answer: C

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